Question

pH of the solution containing $50.0{\text{ mL}}$ of $0.3{\text{ M HCl}}$ and $50.0{\text{ mL}}$ of $0.4{\text{ M N}}{{\text{H}}_3}$ is: $\left[ {{\text{p}}{{\text{K}}_{\text{a}}}\left( {{\text{NH}}_4^ + } \right) = 9.26} \right]$
A) 4.74
B) 9.26
C) 8.78
D) 4.63

Answer 1

Hint: We know that the degree of alkalinity or acidity of a solution is known as its pH. pH is the negative logarithm of the hydrogen ion concentration. We are given hydrochloric acid which is a strong acid and ammonia which is a weak base. To solve this we have to use the equation for the pH of weak acid and strong base.


Complete solution:
We know that the degree of alkalinity or acidity of a solution is known as its pH. pH is the negative logarithm of the hydrogen ion concentration.
We are given that $50.0{\text{ mL}}$ of $0.3{\text{ M HCl}}$ is mixed with $50.0{\text{ mL}}$ of $0.4{\text{ M N}}{{\text{H}}_3}$. Here, ${\text{HCl}}$ is known as hydrochloric acid which is a strong acid and ${\text{N}}{{\text{H}}_3}$ is known as ammonia which is a weak base. The reaction is as follows:
${\text{HCl}} + {\text{N}}{{\text{H}}_3} \to {\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$
Thus, the salt of strong acid and weak base is formed.
First we will calculate the number of moles of ${\text{HCl}}$ and ${\text{N}}{{\text{H}}_3}$ each.
We are given $50.0{\text{ mL}}$ of $0.3{\text{ M HCl}}$. $0.3{\text{ M HCl}}$ means that $0.3{\text{ mol}}$ of ${\text{HCl}}$ are present in $1000{\text{ ml}}$. Thus,
Number of moles of ${\text{HCl}}$ $ = 50{\text{ ml}} \times \dfrac{{0.3{\text{ mol}}}}{{1000{\text{ ml}}}} = 0.015{\text{ mol}}$
We are given $50.0{\text{ mL}}$ of $0.4{\text{ M N}}{{\text{H}}_3}$. $0.4{\text{ M N}}{{\text{H}}_3}$ means that $0.4{\text{ mol}}$ of ${\text{N}}{{\text{H}}_3}$ are present in $1000{\text{ ml}}$. Thus,
Number of moles of ${\text{N}}{{\text{H}}_3}$ $ = 50{\text{ ml}} \times \dfrac{{0.4{\text{ mol}}}}{{1000{\text{ ml}}}} = 0.020{\text{ mol}}$
We can see that there are $0.015{\text{ mol}}$ of ${\text{HCl}}$. From the reaction stoichiometry, we require only $0.015{\text{ mol}}$ of ${\text{N}}{{\text{H}}_3}$. Thus, the remaining amount of ${\text{N}}{{\text{H}}_3}$ is $0.005{\text{ mol}}$.
Thus, in the solution we have $0.005{\text{ mol}}$ of ${\text{N}}{{\text{H}}_3}$ and $0.015{\text{ mol}}$ of ${\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}$.
The equation to calculate the pH is as follows:
\[{\text{pOH}} = {\text{p}}{{\text{K}}_{\text{b}}} + \log \dfrac{{[{\text{salt}}]}}{{[{\text{base}}]}}\]
Where $C$ is the concentration of the salt.
We are given that the ${\text{p}}{{\text{K}}_{\text{a}}}$ for $\left( {{\text{NH}}_4^ + } \right) = 9.26$. Thus,
${\text{p}}{{\text{K}}_{\text{w}}} = {\text{p}}{{\text{K}}_{\text{a}}} + {\text{p}}{{\text{K}}_{\text{b}}}$
${\text{p}}{{\text{K}}_{\text{b}}} = {\text{p}}{{\text{K}}_{\text{w}}} - {\text{p}}{{\text{K}}_{\text{a}}}$
${\text{p}}{{\text{K}}_{\text{b}}} = {\text{14}} - {\text{9}}{\text{.26}}$
${\text{p}}{{\text{K}}_{\text{b}}} = 4.74$
Thus,
\[{\text{pOH}} = {\text{p}}{{\text{K}}_{\text{b}}} + \log \dfrac{{[{\text{salt}}]}}{{[{\text{base}}]}}\]
\[{\text{pOH}} = {\text{4}}{\text{.74}} + \log \dfrac{{0.015{\text{ mol}}}}{{0.005{\text{ mol}}}}\]
\[{\text{pOH}} = {\text{4}}{\text{.74}} + \log \left( 3 \right)\]
\[{\text{pOH}} = {\text{4}}{\text{.74}} + 0.48\]
\[{\text{pOH}} = 5.22\]
Now, calculate the pH using the equation as follows:
${\text{pH}} + {\text{pOH}} = {\text{14}}$
           ${\text{pH}} = {\text{14}} - {\text{pOH}}$
           ${\text{pH}} = {\text{14}} - 5.22$
           ${\text{pH}} = 8.78$
Thus, the pH of a solution containing $50.0{\text{ mL}}$ of $0.3{\text{ M HCl}}$ and $50.0{\text{ mL}}$ of $0.4{\text{ M N}}{{\text{H}}_3}$ is 8.87.

Thus, the correct option is (C) 8.87.


Note: If the pH of the solution is less than 7 then the solution is acidic in nature. If the pH of the solution is equal to 7 then the solution is neutral in nature. If the pH of the solution is more than 7 then the solution is basic in nature. Here, the pH is 8.87 thus, we can say that the solution is basic or alkaline in nature.